Lecture 74 Biweekly: Divide the Array into Pairs of Equal Numbers

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1. Problem description

You are given an array of integers  nums , which contains  2 * n integers.

You need to  nums divide into  n pairs, satisfying:

• Each element  belongs to only one pair.
• Elements in the same pair are  equal  .

If it can be  nums divided into  n pairs, please return  true , otherwise return  false .

Topic link: Divide an array into pairs of equal numbers .

Second, the subject requirements

Sample

输入： nums = [3,2,3,2,2,2]
输出： true
解释：
nums 中总共有 6 个元素，所以它们应该被划分成 6 / 2 = 3 个数对。
nums 可以划分成 (2, 2) ，(3, 3) 和 (2, 2) ，满足所有要求。
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1.简单模拟判断、位运算
2.建议用时10~35min
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3. Problem Analysis

1. Simulation

There are many ways to do this question. At first, I sorted the array from small to large, and then the two are judged as a group.

Don't judge the end of the loop, as long as the two arrays of a group are not equal, just return false.

2. Bit operations

If you don't understand the relevant knowledge points of bit operation, you can read this article, which explains in more detail:

Algorithm questions are practiced daily --- Day 45: Bit operations .

Bit operation can XOR two identical numbers to 0, then we XOR the array from the beginning to the end, if the output is 0, then the condition is met.

But this method has a bug, that is, the result of XOR of unequal numbers may be 0, so the second-to-last example is difficult.

Fourth, the encoding implementation

class Solution {
public:
    bool divideArray(vector<int>& nums) {
        int i,n=nums.size();
        sort(nums.begin(),nums.end());//排序
        for(i=0;i<n-1;i+=2)//两两一组
        {
            if(nums[i]!=nums[i+1])//不相等
                return false;//返回错误
        }
        return true;//返回正确    
    }
};
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5. Test results